intergration (double)

**turion645344** · July 9th 2007, 08:35 AM

evaluate the intergral

3 2
(integral sign) (intergral sign) (2y3+6xy+x2+1)dydx
0 1

so its like 0<x<3 and 1<y<2 (i think)

any help?!

**ThePerfectHacker** · July 9th 2007, 08:48 AM

Originally Posted by turion645344

evaluate the intergral

3 2
(integral sign) (intergral sign) (2y3+6xy+x2+1)dydx
0 1

so its like 0<x<3 and 1<y<2 (i think)

any help?!

$\int_0^3 \int_1^2 2y^3 + 6xy + x^2 +1 \ dy \ dx$

$\int_0^3 \frac{1}{2}y^4+3xy^2+x^2y+y \big|_1^2 \ dx$

$\int_0^3 8+12x+2x^2+2 - \frac{1}{2} - 3x - x^2 - 1 \ dx$

You can do it fron there.

**turion645344** · July 10th 2007, 04:13 AM

when you take them away, i get

3
(intergration sign) x2+9x+8.5 dx

0

when
3
(1/3x3 + 9/2x2 + 8.5x)
0

and then 9+40.5+25.5 so an answer of 75.

**turion645344** · July 11th 2007, 07:48 AM

the answer of 75. is this correct???

**ThePerfectHacker** · July 11th 2007, 09:22 AM

Originally Posted by turion645344

the answer of 75. is this correct???

Yes

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